3/4x^2+1=16

Solution for 3/4x^2+1=16 equation:

3/4x^2+1=16

We move all terms to the left:

3/4x^2+1-(16)=0


Domain of the equation: 4x^2!=0

x^2!=0/4

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

3/4x^2-15=0

We multiply all the terms by the denominator


-15*4x^2+3=0

Wy multiply elements

-60x^2+3=0

a = -60; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-60)·3
Δ = 720
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{5}}{2*-60}=\frac{0-12\sqrt{5}}{-120}
=-\frac{12\sqrt{5}}{-120}
=-\frac{\sqrt{5}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{5}}{2*-60}=\frac{0+12\sqrt{5}}{-120}
=\frac{12\sqrt{5}}{-120}
=\frac{\sqrt{5}}{-10}
$